I wrote this sample KornShell (ksh) code but it is getting bad substitution error during the if clause.
while ((i < $halflen))
do
if [[${strtochk:i:i}==${str
You are using ksh88
but the code you tried is using ksh93
feature missing for the 88 version.
You need to replace
if [[${strtochk:i:i}==${strtochk:j:j}]];then
with these portable lines:
if [ "$(printf "%s" "$strtochk" | cut -c $i)" =
"$(printf "%s" "$strtochk" | cut -c $j)" ]; then
and the incorrect:
i++
j--
with:
i=$((i+1))
j=$((j-1))
shell syntax is very whitespace sensitive:
[[
is acually the name of a command, it's not just syntax, so there must be a space following it. [[
must be ]]
, so it needs to be preceded by a space. [[
works differently depending on the number of arguments it receives, so you want to have spaces around ==
=
.Tips:
break
out of the while loop${strtochk:i:1}
i++
and j--
are arithmetic expressions, not commands, so you need the double parentheses.i=0
and j=$((${#strtochk} - 1))
?while ((i < halflen))
do
if [[ ${strtochk:i:1} == ${strtochk:j:1} ]];then
((i++))
((j--))
else
ispalindrome=false
break
fi
done
Check if your system has rev, then you can simply do:
if [[ $strtochk == $( rev <<< "$strtochk" ) ]]; then
echo "'$strtochk' is a palindrome"
fi
function is_palindrome {
typeset strtochk=$1
typeset -i i=1 j=${#strtochk}
typeset -i half=$(( j%2 == 1 ? j/2+1 : j/2 ))
typeset left right
for (( ; i <= half; i++, j-- )); do
left=$( expr substr "$strtochk" $i 1 )
right=$( expr substr "$strtochk" $j 1 )
[[ $left == $right ]] || return 1
done
return 0
}
if is_palindrome "abc d cba"; then
echo is a palindrome
fi